Fermat 2005 Q20

---

Problem Statement

In triangle $ABC$, if $AB = AC = x + 1$ and $BC = 2x - 2$, where $x > 1$, then the area of the triangle is always equal to

Problem Link

Solution

$$\text{Let the Area of }\triangle ABC \text{ be A }$$ $$\therefore A = \dfrac{1}{2} \times BC \times AD$$ $$A = \dfrac{1}{2} \times (2x - 2) \times \sqrt{((x+1)^2) - ((x-1)^2)}$$ $$A = (x-1) \sqrt{(x^2 + 2x + 1) - (x^2 - 2x + 1)} = (x-1)\sqrt{4x} = 2(x-1)\sqrt{x}$$